# $[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function.

Solve the equation in interval $[0,\pi]:[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function.

How should i start this question,breaking it into intervals is difficult.Please guide me.

• There is the obvious $0$ and $\pi/2$. Nowhere in between. Then can't happen until $3\pi/4$, and then it does. – André Nicolas Sep 20 '15 at 4:53
• Note it does not happen at $3\pi/4$, but afterwards it does. – André Nicolas Sep 20 '15 at 5:00

Key facts: (1) For $0\lt x\lt\pi/2$, we have $\sin x+\cos x\gt 1$, so we do not have equality.

(2) For $\pi/2\lt x\le 3\pi/4$, we have $\lfloor \cos x\rfloor=-1$ but $\lfloor \cos x+\sin x\rfloor=0$.

(3) For $3\pi/4\lt x\le \pi$ we have $\lfloor \cos x\rfloor=-1$ and $\lfloor \cos x+\sin x\rfloor=-1$ so the equality holds.

It remains to check $0$ and $\pi/2$.

Clearly the equality holds true if one of $\sin x,\cos x=0$

This $\implies x=0,\dfrac\pi2,\pi$

$[\sin x]=\begin{cases} 1 &\mbox{if } x=\dfrac\pi2 \\ 0 & \text{otherwise} \end{cases}$

$[\cos x]=\begin{cases} 1 &\mbox{if } x=0 \\ 0 & \mbox{if } 0<x<\dfrac\pi2\\ -1 & \mbox{if } \dfrac\pi2<x\le\pi\\ \end{cases}$

For $0<x<\dfrac\pi2,[\cos x+\sin x]=0\implies0\le\cos x+\sin x<1$

But $\cos x+\sin x=\sqrt2\sin\left(\dfrac\pi4+x\right)$

$\implies0\le\sin\left(\dfrac\pi4+x\right)<\dfrac1{\sqrt2}$

In $0\le x\le\pi$ we need $\dfrac\pi2<x\le\dfrac{3\pi}4$

For $\dfrac\pi2<x<\pi,[\cos x+\sin x]=-1\implies-1\le\cos x+\sin x<0$

$\implies-\dfrac1{\sqrt2}\le\sin\left(\dfrac\pi4+x\right)<0$

In $0\le x\le\pi$ we need $\dfrac{3\pi}4<x\le\pi$

• How is it solved? – Narasimham Sep 20 '15 at 4:50
• @Narasimham, Please find the updated version – lab bhattacharjee Sep 20 '15 at 5:08

Here $x\in \left[0,\pi\right]\;,$ Then $0\leq \sin x\leq 1$. So we get $\displaystyle \lfloor \sin x\rfloor = 0,1$

Similarly $x\in \left[0,\pi\right]\;,$ Then $-1 \leq \cos x\leq 1$. So we get $\displaystyle \lfloor \cos x\rfloor = -1, 0,1$

Similarly $x\in \left[0,\pi\right]\;,$ Then $-1 \leq \sin x+ \cos x\leq \sqrt{2}$. So we get $\displaystyle \lfloor \sin x+\cos x\rfloor =-1, 0,1$

$\bullet \;$ If $\displaystyle \lfloor \sin x+\cos x\rfloor=0\;,$ Then $\lfloor \sin x \rfloor =0$ and $\lfloor \cos x \rfloor =0$

$\bullet \;$ If $\displaystyle \lfloor \sin x+\cos x\rfloor=0\;,$ Then $\lfloor \sin x \rfloor =1$ and $\lfloor \cos x \rfloor =-1$

$\bullet \;$ If $\displaystyle \lfloor \sin x+\cos x\rfloor=1\;,$ Then $\lfloor \sin x \rfloor =0$ and $\lfloor \cos x \rfloor =1$

$\bullet \;$ If $\displaystyle \lfloor \sin x+\cos x\rfloor=1\;,$ Then $\lfloor \sin x \rfloor =1$ and $\lfloor \cos x \rfloor =0$

$\bullet \;$ If $\displaystyle \lfloor \sin x+\cos x\rfloor=-1\;,$ Then $\lfloor \sin x \rfloor =0$ and $\lfloor \cos x \rfloor =-1$